Factoring and graph isomorphism are problems in NP that are not known to be in P nor to be NP-Complete. What are some other (sufficiently different) natural problems that share this property? Artificial examples coming directly from the proof of Ladner's theorem do not count.

Are any of these example provably NP-intermediate, assuming only some "reasonable" hypothesis?

There are several complexity classes between P and NP-complete which are currently regarded as interesting: PPAD, problems that are UGC-equivalent, NP $\cap$ co-NP, BPP,.... If you are asking for a big list, could you make this a community wiki please?
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András SalamonAug 17 '10 at 2:36

Thank you. I am aware of Ladner's Theorem. I guess I was asking for "natural problems." I guess PPAD has Nash Equilibria, so that counts...
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Lev Reyzin♦Aug 17 '10 at 3:05

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Lev, this should be community wiki, since there is no single right answer. Please flag it as such.
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MoritzAug 17 '10 at 3:17

Are any of these examples provably NP-intermediate, assuming only some "reasonable" hypothesis (i.e., a hypothesis less trivial than "this problem is NP-intermediate")? If so, it would be interesting to mention that in this list.
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Timothy ChowJun 25 '11 at 17:57

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@Timothy Chow, @Lev Reyzin. Graph isomorphism is not NP-complete unless $\Sigma_2^p = \Pi_2^p$, because graph non-isomorphism is in AM, and coNP in AM implies $\Sigma_2^p = \Pi_2^p$. This would work for other problems whose complements are in AM. This of course does not rule out GI being in P.
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Jeff KInneJun 30 '11 at 19:59

My favorite problem in this class (I'll phrase it as a functional problem, but it's easy to turn into a decision problem in the standard way): compute the rotation distance between two binary trees (equivalently, the flip distance between two triangulations of a convex polygon).

That's a neat problem: I didn't realize it was in limbo.
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Suresh Venkat♦Aug 17 '10 at 3:29

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Yeah I didn't know about it either! For all these problems/answers, I wonder if they are in Limbo because we think they really are or if they're more like PRIMES...
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Lev Reyzin♦Aug 17 '10 at 5:21

This problem and its potentially intermediate status should be more well-known. Can you give a reference to it? Also, is there any result indicating that it is not NP-complete, as there is for Graph Isomorphism and related problems?
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Joshua GrochowAug 17 '10 at 5:32

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A very pretty and important but older reference is Thurston, Sleator, and Tarjan, "Rotation distance, triangulations, and hyperbolic geometry", STOC'86 and JAMS'88. For a recent reference that explicitly mentions the complexity of the problem as still being open, see Lucas, "An improved kernel size for rotation distance in binary trees", IPL 2010, dx.doi.org/10.1016/j.ipl.2010.04.022
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David EppsteinAug 17 '10 at 6:01

A problem that is mentioned neither in this list or the MO list is the turnpike problem. Given a multiset of n(n-1)/2 numbers, each number representing the distance between two points on the line, reconstruct the positions of the original points.

Note that what makes this nontrivial is that for a given number d in the multiset, you don't know which pair of points is d units apart.

While it is known that for any given instance there are only a polynomial number of solutions, it is not known how to find one !

Thanks - this is a good one! Reminds me of some other "localization" problems. Is it actually thought to be not in p?
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Lev Reyzin♦Aug 17 '10 at 5:18

I am not aware that turnpike is directly linked to known problems in complexity. However, there's a "wrong direction" relation to factoring, in that the turnpike problem cam be phrased as a factoring problem on an appropriately chosen polynomial.
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Suresh Venkat♦Aug 17 '10 at 5:22

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Are there known unlikely consequences of this problem being NP-complete, as there are for Graph Isomorphism (PH collapsing)?
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Joshua GrochowAug 17 '10 at 5:37

not that I'm aware of. it hasn't been studied that much, which is a pity, because it's so natural.
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Suresh Venkat♦Aug 17 '10 at 5:46

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You encounter a similar problem in bioinformatics: Given a set of potentially/hopefully overlapping, randomly created substrings of a string much longer than the individual pieces; compute the original string. (gene sequencing)
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RaphaelOct 26 '10 at 22:02

Here is a list of problems that may or may not qualify as "sufficiently" different. By the same proof as for Graph Isomorphism, if any of them is NP-complete, then the Polynomial Hierarchy collapses to the second level. I do not think there is any broad consensus as to which of these "ought" to be in P.

Graph Automorphism (determine if a graph has a nontrivial automorphism). Reduces to Graph Isomorphism, but not known (not thought?) to be GI-hard.

Group Isomorphism and Automorphism (where the groups are given by their multiplication tables). Again, reduces to Graph Isomorphism, but not thought to be GI-hard.

This post by Bill Gasarch contains some other problems with the taste of Ramsey theory that look like they might be intermediate.

By Mahaney's Theorem, no sparse set can be NP-complete. But we also know that there are sparse sets in $NP$ - $P$ iff $NEXP$ is not equal to $EXP$. So assuming $NEXP \neq EXP$, the padded version of any $NEXP$-complete problem is of intermediate complexity. (Such a set cannot be in $P$ unless $NEXP = EXP$, contradicting our assumption.) There are plenty of natural $NEXP$-complete problems.

The Pythogorean theorem implies that the length of any polygonal curve whose vertices and integer endpoints is a sum of square roots of integers. Thus, the sum-of-roots problem is inherent in several planar computational geometry problems, including Euclidean minimum spanning trees, Euclidean shortest paths, minimum-weight triangulations, and the Euclidean traveling salesman problem. (The Euclidean MST problem can be solved in polynomial time without resolving the sum-of-roots problem, thanks to the underlying matroid structure and the fact that the EMST is a subgraph of the Delaunay triangulation.)

There is a polynomial-time randomized algorithm, due to Johannes Blömer, to decide whether the two sums are equal. However, if the answer is no, Blömer's algorithm does not determine which sum is larger.

The decision version of this problem (Is $A > B$?) is not even known to be in NP. However, Blömer's algorithm implies that if the decision problem is in NP, then it is also in co-NP. Thus, the problem is unlikely to be NP-complete.

The Minimum Circuit Size Problem (MCSP) is my favorite "natural" problem in NP that's not known to be NP-complete: Given the truth-table (of size n=2^m) of an m-variate Boolean function f, and given a number s, does f have a circuit of size s? If MCSP is easy, then there is no cryptographically-secure one-way function. This problem and its variants provided much of the motivation for the study of "brute-force" algorithms in Russia, leading to Levin's work on NP-completeness. This problem can also be viewed in terms of resource-bounded Kolmogorov complexity: asking if a string can be recovered quickly from a short description. This version of the problem was studied by Ko; the name MCSP was used first by Cai and Kabanets, as far as I know. More references can be found in some papers of mine:
http://ftp.cs.rutgers.edu/pub/allender/KT.pdf
http://ftp.cs.rutgers.edu/pub/allender/pervasive.reach.pdf

It might be worth mentioning that, as with many potentially NP-intermediate problems, a slight variant is known to be NP-complete. Namely, 3-manifold knot genus is NP-complete: given a closed polygonal chain in a triangulated 3-manifold and an integer g, is the knot the boundary of a surface of genus at most g? (Being the unknot is equivalent to genus 0.) doi.acm.org.proxy.uchicago.edu/10.1145/509907.510016
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Joshua GrochowOct 25 '10 at 2:45

This problem is in co-NP[$\log^2 n$], i.e., it is decidable with $O(\log^2n/ \log\log n)$ nondeterministic steps. Thus, it has a quasi-polynomial time algorithm ($O(n^{\log n/\log \log n})$ time) so it is unlikely to be co-NP-hard.

It is not known if it is possible to decide in polynomial time if player 1 has a winning strategy in a parity game (from a given starting position). The problem is, however, contained in NP and co-NP and even in UP and co-UP.

M. Jurdzinski. Deciding the Winner in Parity Games is in UP \cap co-Up. Information Processing Letters 68(3):119-124. 1998. Should at least be a good starting point.
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MatthiasAug 17 '10 at 16:43

The recent paper "A Pumping Algorithm for Ergodic Stochastic Mean Payoff Games with Perfect Information" also shows that even a generalization of the parity game can be solved in pseudo-polynomial time. In particular, they show that a game called BWR game has a pseudo-polynomial time algorithm when there is a constant number of "random nodes". The parity game is the case where there is no random nodes.
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DanuDec 23 '10 at 0:52

You get a very long list of problems if willing to accept approximation problems, such as approximating Max-Cut to within a factor 0.878. We don't know if it's NP-hard or in P (only know NP-hardness assuming the Uniuqe Games Conjecture).

In a monotone CNF formula every clause contains only positive literals or only negative literals. In an intersecting monotone CNF formula every positive clause has some variable in common with every negative clause.

Here's a problem in computational social choice which is not known to be in P, and may or may not be NP-complete.

Agenda control for balanced single-elimination tournaments:

Given: tournament graph $T$ on $n=2^k$ nodes, node $a$

Question: does there exist a permutation of the nodes (a bracket) so that a is the winner of the induced single-elimination tournament?

Given a permutation $P_k$ on $2^k$ nodes of $V$ and a tournament graph $T$ on $V$, one can obtain a permutation $P_{k-1}$ on $2^{k-1}$ nodes as follows. For every $i>0$, consider $P_k[2i-1]$ and $P_k[2i]$ and the arc $e$ between them in $T$; let $P_{k-1}[i]=P_k[2i-1]$ if $e=(P_k[2i-1],P_k[2i])$ and $P_{k-1}[i]=P_k[2i]$ otherwise.
That is, we match up pairs of nodes according to $P_k$ and use $T$ to decide which nodes (winners) move on to the next round $P_{k-1}$. Hence given a permutation on $2^k$ one can actually define $k$ rounds $P_{k-1},\ldots,P_0$ inductively as above, until the last permutation contains only one node. This defines a (balanced) single-elimination tournament on $2^k$ nodes. The node which remains after all the rounds is the winner of the tournament.

A binomial arborescence on $2^k$ nodes rooted at a node $x$ is defined recursively as $a$ binomial arborescence on $2^{k-1}$ nodes rooted at $x$ and a binomial arborescence on $2^{k-1}$ nodes rooted at a different node $y$ and an arc from $x$ to $y$. (If $k=0$, a binomial arborescence is just the root.) The spanning binomial arborescences in a tournament graph capture exactly the single-elimination tournaments which can be played, given the match outcome information in the tournament graph.

Although this is an interesting result, if you check the paper it even says that the proof of intermediate complexity is essentially the same as Ladner's Theorem, except you do the diagonalization in the choice of the LHS restriction. So I don't know if this counts as a "natural" problem, rather than just a different encoding of Ladner's Theorem.
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Joshua GrochowAug 19 '10 at 2:50

Note also that these are source-and-target restrictions. The target (right hand side) has to be of special form, to enforce injectivity.
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András SalamonSep 22 '10 at 14:59

A graph $G = (V,E)$ is said to be labeled by $f$ if each vertex $v\in V$ is assigned a non-negative integer value $f(v)$ and each edge $e = uv\in E$ is assigned the value $|f(u) − f(v)|$. The labeling is graceful if $f : V\rightarrow \{0, 1, 2,\dots, |E|\}$ is an injection and if all edges of G have distinct labels from $\{1, 2,..., |E|\}$. A graph is said to be graceful if it admits a graceful labeling. The computational complexity of deciding whether a graph is graceful is not known.

The gap version of the closest vector in lattice problem is the following: Given a basis for an $n$-dimensional lattice and a vector $v$, distinguish between the two cases where there is a lattice vector at distance at most one $1$ from $v$ or when every lattice vector is $\beta$-far from $v$, for some fixed gap parameter $\beta > 1$.

When $\beta = \sqrt{n}$, the problem is in $\mathsf{NP} \cap \mathsf{coNP}$ and thus unlikely to be $\mathsf{NP}$-complete (and is conjectured to be outside $\mathsf{P}$). This case is of central attention for lattice-based cryptography and the related dihedral hidden subgroup problem in quentum computing. When $\beta$ is much smaller, say $\beta = n^{o(1/\log \log n)}$, the problem becomes $\mathsf{NP}$-hard.

In the linear divisibility problem, the input is two integers $a$ and $b$ and the task is to determine whether there exists an integer of the form $a \cdot x+1$ that divides $b$.

The linear divisibility problem is known to be $\gamma$-complete for NP but not known (AFAIK) to be NP-complete.

Garey and Johnson in their seminal "Computers and Intractability" say that (pp. 158-159):

A $\gamma$-reduction, in contrast to our other notions of reducibility, is nondeterministic in nature. Let us first introduce the notion of the relation $R_M$ computed by an NDTM program $M$:$$R_M=\{\langle x,y\rangle:there\ is\ a\ string\ z\ such\ that\ on\ input\ x\ and\ guess\ z\ M\ has\ output\ y\}$$
(where the definition of "output" is as in the computation of functions by DTMs).

We say that a language $L_1$ over alphabet $\Sigma_1$ is $\gamma$-reducible to a language $L_2$ over $\Sigma_2$ (written $L_1\propto_{\gamma}L_2$) if there is a polynomial time NDTM program $M$ such that for all $x\in\Sigma_1^*$ there is some $y\in\Sigma_2^*$ for which $\langle x,y\rangle\in R_M$ and such that for all $\langle x,y\rangle\in R_M$, $x\in L_1$ if and only if $y\in L_2$. In other words, there is at least one halting computation for $M$ on every input $x$ and, given an input $x$, all halting computations on $x$ yield outputs that are in $L_2$ if and only if $x\in L_1$.

The problem of finding Vapnik–Chervonenkis dimension is not known to be in $P$ neither known to be $NP$-complete. The problem can be solved by quasi-polynomial time algorithm ($ O(n^{\log n})$). The problem can not be $NP$-complete unless $NP$ is contained in Quasi-polynomial time.

I think it's worth mentioning the classes $\mathsf{LOGNP}$ and $\mathsf{NP}[\log^2 n]$ defined in the paper linked in the answer as well as the other natural problems in that paper that are shown to be in these classes, since any problem complete for these classes is likely to be of intermediate complexity.
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Joshua GrochowNov 21 '13 at 17:58

The following problem is believed to be NP-Intermediate, i.e. it is in NP but neither in P nor NP-complete.

Exponentiating Polynomial Root Problem (EPRP)

Let $p(x)$ be a polynomial with $\deg(p) \geq 0$ with coefficients drawn from a finite field $GF(q)$ with $q$ a prime number, and $r$ a primitive root for that field. Determine the solutions of:
$$p(x) = r^x $$
(or equivalently, the zeros of $p(x) - r^x$) where $r^x$ means exponentiating $r$.

Note that, when $\deg(p)=0$ (the polynomial is a constant), this problem reverts to the Discrete Logarithm Problem, which is believed to be NP-Intermediate as well.

The problem of finding a minimum dominating set in tournament is not known to be in $P$ neither known to be $NP$-complete. The problem has quasi-polynomial time algorithm ($ O(n^{\log n})$). If the problem has polynomial time algorithm then satisfiability can be solved in sub-exponential time.